Question

Solve the following rational equations.

Answer 1

\[\frac{z}{6+z}+\frac{z-1}{6-z}=\frac{z}{6-z}\]

Answer 2

z=3
z=2

Answer 3

We have
\[\frac{z}{6+z} +\frac{z-1}{6-z}=\frac{z}{6-z}\]
Let's take the first two terms, the LCM of (6+z) and (6-z) is (6+z)(6-z)
so we get
\[\frac{z(6-z)}{(6+z)(6-z)} +\frac{(z-1)(6+z)}{(6-z)(6+z)}=\frac{z}{6-z}\]
Let's expand the numerator on the left hand side
\[\frac{6z-z^2}{(6+z)(6-z)} +\frac{(6z+z^2-6-z)}{(6-z)(6+z)}=\frac{z}{6-z}\]
Let's combine the like terms now
\[\frac{6z-z^2+6z+z^2-6-z}{(6+z)(6-z)} =\frac{z}{6-z}\]
we get now
\[\frac{11z-6}{(6+z)(6-z)} =\frac{z}{6-z}\]
now let's cancel the common term from both side
\[\frac{11z-6}{(6+z)\cancel {(6-z)}} =\frac{z}{\cancel {(6-z)}}\]
cross multiplying we get
\[11z-6=6z+z^2\]
Let's take all the terms to one side, and combine like terms
\[z^2-5z+6=0\]
Let's find factors of 6 which have a sum of -5,
which are -3 and -2
\[z^2-2z-3z+6=0\]
now take z common from first two terms and -3 from last two terms
\[z(z-2)-3(z-2)=0\]
we get
\[(z-3)(z-2)=0\]
we get z=3 or 2